Q. 135.0( 1 Vote )

# Two cards are sel

As box contains cards numbered as 1,1,2,2 and 3

possible sums of card numbers are 2,3,4 and 5

Hence, X can take values 2,3,4 and 5

X=2 [ when drawn cards are (1,1)]

X=3 [when drawn cards are (1,2) or (2,1)]

X=4 [when drawn cards are (2,2) or (3,1) or (1,3)]

X=5 [when drawn cards are (2,3) or (3,2)]

As Y is a random variable representing maximum of the two numbers drawn

Y can take values 1,2 and 3.

Y=1 [when drawn cards are 1 and 1]

Y=2 [when drawn cards are (1,2) or (2,2) or (2,1)]

Y=3 [when drawn cards are (1,3) or (3,1) or (2,3) or (3,2)]

Note : P(1) represents probability of drawing card numbered as 1, similarly P(2) and P(3)

P(X=2) = P(1)P(1) =

[For drawing first card we had 2 favourable outcomes as 1,1 out of total 5 ,in second time of drawing ,as we drew a card numbered as 1 we are having 1 favourable outcome out of total remaining of 4]

Similarly,

P(X=3) = P(2)P(1) + P(1)P(2) =

P(X=4) = P(2)P(2)+P(3)P(1)+P(1)P(3) =

P(X=5) = P(2)P(3)+P(3)P(2) =

Similarly,

P(Y=1) = P(1)P(1) =

P(Y=2) = P(1)P(2)+P(2)P(1)+P(2)P(2) =

P(Y=3) = P(2)P(3)+P(3)P(2)+ P(1)P(3)+P(3)P(1)

=

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi2pi – (∑xipi)2

first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

Mean for(X) = 0.2+1.2+1.2+1 = 3.6

Variance for(X) = 0.4+3.6+4.8+5.0-3.62 = 13.8-3.62 = 0.84

Similarly probability distribution for Y is given below:

Mean for(Y) = 0.1+1.0+1.2 = 2.3

Variance for(Y) = 0.1+2.0+3.6-2.32 = 5.7-2.32 = 0.41

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